S2T2

Jim Randell 8:11 am on 6 March 2026 Permalink Reply
Tags: by: Andrew Skidmore ( 86 )   

• 

  Jim Randell 8:13 am on 6 March 2026 Permalink | Reply

  Here is a solution using the [[ SubstitutedExpression ]] solver from the enigma.py library, that is a direct interpretation of the puzzle text.

  The following run file executes in 86ms. (Internal runtime of the generated code is 5.8ms).

  #! python3 -m enigma -rr
  
  SubstitutedExpression
  
  "SIX * TWO = TWELVE"
  "FIVE % 5 = 0"
  "SIX % 6 = 0"
  
  --answer="SOLVE"
  

  Solution: SOLVE = 95380.

  We have:

  TWO = 165
  FIVE = 4780
  SIX = 972
  TWELVE = 160380

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• 

  ruudvanderham 2:45 pm on 6 March 2026 Permalink | Reply

  Very brute force:

  for s, i, x, t, w, o, l, f, e, v in istr.permutations(range(10)):
      if (s | i | x) * (t | w | o) == (t | w | e | l | v | e) and (f | i | v | e).is_divisible_by(5) and (s | i | x).is_divisible_by(6):
          peek(s | o | l | v | e)
  

  and a bit less brute:

  for five in istr.range(1000, 10000, 5):
      for six in istr.range(102, 1000, 6):
          if five[1] == six[1]:
              istr.decompose(five | six, "fivesix")
              if len(fivesix := set(five) | set(six)) == 6:
                  for t, w, o in istr.permutations(set(istr.range(10)) - set(fivesix), 3):
                      twelve = six * (two := t | w | o)
                      if len(twelve) == 6:
                          istr.decompose(twelve, "TWElVF")
                          if T == t and W == w and E == F == e and V == v:
                              if len({f, i, v, e, s, i, x, t, w, o, t, w, e, l, v, e}) == 10:
                                  solve = istr("=solve")
                                  peek(five, six, two, twelve, solve)
  

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